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# The Real and Complex Numbers: Properties of the Real Numbers

Module by: Lawrence Baggett. E-mail the author

Summary: Properties of the real numbers, discussing isomorphic subsets, nonempty subsets with a greatest lower bound, least upper bound properties, positive square roots, and other aspects of real numbers.

## Theorem 1

The set RR contains a subset that is isomorphic to the ordered field QQ of rational numbers, and hence subsets that are isomorphic to NN and Z.Z.

REMARK. The proof of Statement 1 is immediate from part (b) of Exercise 1.7. In view of this theorem, we will simply think of the natural numbers, the integers, and the rational numbers as subsets of the real numbers.

Having made a definition of the set of real numbers, it is incumbent upon us now to verify that this set RR satisfies our intuitive notions about the reals. Indeed, we will show that 22 is an element of RR and hence is a real number (as plane geometry indicates it should be), and we will show in later chapters that there are elements of RR that agree with our intuition about ee and π.π. Before we can proceed to these tasks, we must establish some special properties of the field R.R. The first, the next theorem, is simply an analog for lower bounds of the least upper bound condition that comes from the completeness property.

## Theorem 2

If SS is a nonempty subset of RR that is bounded below, then there exists a greatest lower bound for S.S.

### Proof

Define TT to be the set of all real numbers xx for which -xS.-xS. That is, TT is the set -S.-S. We claim first that TT is bounded above. Thus, let mm be a lower bound for the set S,S, and let us show that the number -m-m is an upper bound for T.T. If xT,xT, then -xS.-xS. So, m-x,m-x, implying that -mx.-mx. Since this is true for all xT,xT, the number -m-m is an upper bound for T.T.

Now, by the completeness assumption, TT has a least upper bound M0.M0. We claim that the number -M0-M0 is the greatest lower bound for S.S. To prove this, we must check two things. First, we must show that -M0-M0 is a lower bound for S.S. Thus, let yy be an element of S.S. Then -yT,-yT, and therefore -yM0.-yM0. Hence, -M0y,-M0y, showing that -M0-M0 is a lower bound for S.S.

Finally, we must show that -M0-M0 is the greatest lower bound for S.S. Thus, let mm be a lower bound for S.S. We saw above that this implies that -m-m is an upper bound for T.T. Hence, because M0M0 is the least upper bound for T,T, we have that -mM0,-mM0, implying that m-M0,m-M0, and this proves that -M0-M0 is the infimum of the set S.S.

The following is the most basic and frequently used property of least upper bounds. It is our first glimpse of “ limits.” Though the argument is remarkably short and sweet, it will provide the mechanism for many of our later proofs, so master this one.

## Theorem 3

Let SS be a nonempty subset of RR that is bounded above, and Let M0M0 denote the least upper bound of S;S; i.e., M0=supS.M0=supS. Then, for any positive real number ϵϵ there exists an element tt of SS such that t>M0-ϵ.t>M0-ϵ.

### Proof

Let ϵ>0ϵ>0 be given. Since M0-ϵ<M0,M0-ϵ<M0, it must be that M0-ϵM0-ϵ is not an upper bound for S.S. (M0M0 is necessarily less than or equal to any other upper bound of S.S.) Therefore, there exists an element tStS for which t>M0-ϵ.t>M0-ϵ. This is exactly what the theorem asserts.

## Exercise 1

1. Let SS be a nonempty subset of RR which is bounded below, and let m0m0 denote the infimum of S.S. Prove that, for every positive δ,δ, there exists an element ss of SS such that s<m0+δ.s<m0+δ. Mimic the proof to Theorem 3.
2. Let SS be any bounded subset of R,R, and write -S-S for the set of negatives of the elements of S.S. Prove that sup(-S)=-infS.sup(-S)=-infS.
3. Use part (b) to give an alternate proof of part (a) by using Theorem 3 and a minus sign.

## Exercise 2

1. Let SS be the set of all real numbers xx for which x<1.x<1. Give an example of an upper bound for S.S. What is the least upper bound of S?S? Is supSsupS an element of S?S?
2. Let SS be the set of all xRxR for which x24.x24. Give an example of an upper bound for S.S. What is the least upper bound of S?S? Does supSsupS belong to S?S?

We show now that RR contains elements other than the rational numbers in Q.Q. Of course this holds for any complete ordered field. The next theorem makes this quite explicit.

## Theorem 4

If xx is a positive real number, then there exists a positive real number yy such that y2=x.y2=x. That is, every positive real number xx has a positive square root in R.R. Moreover, there is only one positive square root of x.x.

### Proof

Let SS be the set of positive real numbers tt for which t2x.t2x. Then SS is nonempty Indeed, If x>1,x>1, then 1 is in SS because 12=1×1<1×x=x.12=1×1<1×x=x. And, if x1,x1, then xx itself is in SS, because x2=x×x1×x=x.x2=x×x1×x=x.

Also, SS is bounded above. In fact, the number 1+x/21+x/2 is an upper bound of S.S. Indeed, arguing by contradiction, suppose there were a tt in SS such that t>1+x/2.t>1+x/2. Then

x t 2 > ( 1 + x / 2 ) 2 = 1 + x + x 2 / 4 > x , x t 2 > ( 1 + x / 2 ) 2 = 1 + x + x 2 / 4 > x ,
(1)

which is a contradiction. Therefore, 1+x/21+x/2 is an upper bound of S,S, and so SS is bounded above.

Now let y=supS.y=supS. We wish to show that y2=x.y2=x. We show first that y2x,y2x, and then we will show that y2x.y2x. It will then follow from the tricotomy law that y2=x.y2=x. We prove both these inequalities by contradiction.

So, assume first that y2>x,y2>x, and write αα for the positive number y2-x.y2-x. Let ϵϵ be the positive number α/(2y),α/(2y), and, using Theorem 1.5, choose a tStS such that t>y-ϵ.t>y-ϵ. Then y+t(2y),y+t(2y), and y-t<ϵ=α/2y.y-t<ϵ=α/2y. So,

α = y 2 - x = y 2 - t 2 + t 2 - x y 2 - t 2 = ( y + t ) ( y - t ) 2 y ( y - t ) < 2 y ϵ < 2 y × α 2 y = α , α = y 2 - x = y 2 - t 2 + t 2 - x y 2 - t 2 = ( y + t ) ( y - t ) 2 y ( y - t ) < 2 y ϵ < 2 y × α 2 y = α ,

which is a contradiction. Therefore y2y2 is not greater than x.x.

Now we show that y2y2 is not less than x.x. Again, arguing by contradiction, suppose it is, and let ϵϵ be the positive number x-y2.x-y2. Choose a positive number δδ that is less than yy and also less than ϵ/(3y).ϵ/(3y). Let s=y+δ.s=y+δ. Then ss is not in S,S, whence s2>x,s2>x, so that we must have

ϵ = x - y 2 = x - s 2 + s 2 - y 2 s 2 - y 2 = ( s + y ) ( s - y ) = ( 2 y + δ ) δ < 3 y δ < ϵ , ϵ = x - y 2 = x - s 2 + s 2 - y 2 s 2 - y 2 = ( s + y ) ( s - y ) = ( 2 y + δ ) δ < 3 y δ < ϵ ,

This completes the proof that y2=x;y2=x; i.e., that xx has a positive square root.

Finally, if y'y' were another positive number for which y'2=x,y'2=x, we show that y=y'y=y' by ruling out the other two cases: y<y'y<y' and y>y'.y>y'. For instance, if y<y',y<y', then we would have that y2<y'2,y2<y'2, giving that

x = y 2 < y ' 2 = x , x = y 2 < y ' 2 = x ,

implying that x<x,x<x, and this is a contradiction.

Definition 1:

If xx is a positive real number, then the symbol xx will denote the unique positive number yy for which y2=x.y2=x. Of course, 00 denotes the number 0.

REMARK Part (c) of (Reference) shows that the field QQ contains no number whose square is 2, and Theorem 4 shows that the field RR does contain a number whose square is 2. We have therefore “proved” that the real numbers is a larger set than the rational numbers. It may come as a surprise to learn that we only now have been able to prove that. Look back through the chapter to be sure. It follows also that QQ itself is not a complete ordered field. If it were, it would be isomorphic to R,R, by Theorem 1.2, so that it would have to contain a square root of 2, which it does not.

Definition 2:

A real number xx that is not a rational number, i.e., is not an element of the subset QQ of R,R, is called an irrational number.

## Exercise 3

1. Prove that every positive real number has exactly 2 square roots, one positive (xx) and the other negative (-x-x).
2. Prove that if xx is a negative real number, then there is no real number yy such that y2=x.y2=x.
3. Prove that the product of a nonzero rational number and an arbitrary irrational number must be irrational. Show by example that the sum and product of irrational numbers can be rational.

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